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An ancient optics problem of Ptolemy, studied later by Alhazen, is discussed. This problem deals with reﬂection of light in spherical mirrors. Mathematically, this reduces to the solution of a quartic equation, which we solve and analyze using a symbolic computation software. Similar problems have been recently studied in connection with ray-tracing, catadioptric optics, scattering of electromagnetic waves, and mathematical billiards, but we were led to this problem in our study of the so-called triangular ratio metric. Keywords Triangular ratio metric · Ptolemy–Alhazen problem · Reﬂection of light Mathematics Subject Classiﬁcation 30C20 · 30C15 · 51M99 1 Introduction The Greek mathematician Ptolemy (ca. 100–170) formulated a problem concerning reﬂection of light at a spherical mirror surface: Given a light source and a spherical Communicated by Stephan Ruscheweyh. B Parisa Hariri parisa.hariri@utu.ﬁ Masayo Fujimura masayo@nda.ac.jp Marcelina Mocanu mmocanu@ub.ro Matti Vuorinen vuorinen@utu.ﬁ Department of Mathematics, National Defense Academy of Japan, Yokosuka, Japan Department of Mathematics and Statistics, University of Turku, Turku, Finland Department of Mathematics and Informatics, Vasile Alecsandri University of Bacau, Bacau, ˘ Romania 123 136 M. Fujimura et al. mirror, ﬁnd the point on the mirror where the light will be reﬂected to the eye of an observer. Alhazen (ca. 965–1040) was a scientist who lived in Iraq, Spain, and Egypt and extensively studied several branches of science. For instance, he wrote seven books about optics and studied, e.g., Ptolemy’s problem as well as many other problems of optics and is considered to be one of the greatest researchers of optics before Kepler [2]. Often the above problem is known as Alhazen’s problem [10, p. 1010]. At the end of this introduction, we will point out various applications and earlier results connected with the Ptolemy–Alhazen problem. We will consider the two-dimensional version of the problem and present an alge- braic solution for it. The solution reduces to a quartic equation which we solve with symbolic computation software. Let D be the unit disk {z ∈ C :|z| < 1}, and suppose that the circumference ∂ D ={z ∈ C :|z|= 1} is a reﬂecting curve. This two-dimensional problem reads: Given two points z , z ∈ D , ﬁnd u ∈ ∂ D such that 1 2 (z , u, 0) = (0, u, z ). (1.1) 1 2 Here, (z, u,w) denotes the radian measure in (−π, π ] of the oriented angle with initial side [u, z] and ﬁnal side [u,w]. This equality condition for the angles says that the angles of incidence and reﬂection are equal, a light ray from z to u is reﬂected at u and goes through the point z . Recall that, according to Fermat’s principle, light travels between two points along a path of extremal time, as compared to other nearby it paths. One proves that u = e , t ∈ R satisﬁes (1.1) if and only if t is a critical 0 0 it it point of the function t → z − e + z − e , t ∈ R. In particular, condition (1.1) 1 2 is satisﬁed by the extremum points (a minimum point and a maximum point, at least) of the function u → |z − u| + |z − u|, u ∈ ∂ D . 1 2 We call this the interior problem—there is a natural counterpart of this problem for the case when both points are in the exterior of the closed unit disk, called the exterior problem. Indeed, this exterior problem corresponds to Ptolemy’s questions about light source, spherical mirror, and observer. As we will see below, the interior problem is equivalent to ﬁnding the maximal ellipse with foci at z , z contained in 1 2 the unit disk, and the point of reﬂection u ∈ ∂ D is the tangent point of the ellipse with the circumference. Algebraically, this leads to the solution of a quartic equation as we will see below. We met this problem in a different context, in the study of the triangular ratio metric s of a given domain G ⊂ R deﬁned as follows for z , z ∈ G [6,12] G 1 2 |z − z | 1 2 s (z , z ) = sup . (1.2) G 1 2 |z − z|+|z − z | 1 2 z∈∂G By compactness, this supremum is attained at some point z ∈ ∂G. If G is convex, it is simple to see that z is the point of contact of the boundary with an ellipse, with foci z , z , contained in G. Now for the case G = D and z , z ∈ D, if the extremal 1 2 1 2 123 The Ptolemy–Alhazen Problem... 137 point is z ∈ ∂ D, the connection between the triangular ratio distance |z − z | 1 2 s (z , z ) = D 1 2 |z − z |+|z − z | 1 0 2 0 and the Ptolemy–Alhazen interior problem is clear: u=z satisﬁes (1.1). Note that (1.1) is just a reformulation of a basic property of the ellipse with foci z , z : the normal to 1 2 the ellipse (which in this case is the radius of the unit circle terminating at the point u) bisects the angle formed by segments joining the foci z , z with the point u.During 1 2 the past decade, the s metric has been studied in several papers e.g. by P. Hästö [13,14]; the interested reader is referred to [12] and the references there. We study the Ptolemy–Alhazen interior problem and in our main result, Theo- rem 1.1, we give an equation of degree four that yields the reﬂection point on the unit circle. Standard symbolic computation software can then be used to ﬁnd this point numerically. We also study the Ptolemy–Alhazen exterior problem. Theorem 1.1 The point u in (1.1) is given as a solution of the equation 4 3 z z u − (z + z )u + (z + z )u − z z = 0. (1.3) 1 2 1 2 1 2 1 2 It should be noted that the Eq. (1.3) may have roots in the complex plane that are not on the unit circle, and of the roots on the unit circle, we must choose one root u , that minimizes the sum |z − u|+|z − u| . We call this root the minimizing root of 1 2 (1.3). Corollary 1.2 For z , z ∈ D we have 1 2 |z − z | 1 2 s (z , z ) = D 1 2 |z − u|+|z − u| 1 2 where u ∈ ∂ D is the minimizing root of (1.3). As we will see below, the minimizing root need not be unique. We have used Risa/Asir symbolic computation software [20] in the proofs of our results. We give a short Mathematica code for the computation of s (z , z ). D 1 2 Theorem 1.1 is applicable not merely to light signals but whenever the angles of incidence and reﬂection of a wave or signal are equal, for instance, in the case of electromagnetic signals like radar signals or acoustic waves. H. Bach [4] has made numerical studies of Alhazen’s ray-tracing problem related to circles and ellipses. A.R. Miller and E. Vegh [18] have studied the exterior Ptolemy–Alhazen problem and computed the grazing angle of specular reﬂection (the complement of the equal angles of incidence and of reﬂection) using a quartic equation, which is not the same as (1.3). They did not consider the problem of ﬁnding the point of incidence in the case of specular reﬂection, which is solved through Eq. (1.3). Mathematical theory of billiards also leads to similar studies: see for instance the paper by M. Drexler and M.J. Gander [9]. The Ptolemy–Alhazen problem also occurs in computer graphics and catadioptric optics [1]. The well-known lithograph of M. C. 123 138 M. Fujimura et al. Escher named “Hand with reﬂecting sphere” demonstrates nicely the idea of reﬂection from a spherical mirror. 2 Algebraic solution to the Ptolemy–Alhazen problem In this section, we prove Theorem 1.1 and give an algorithm for computing s (z , z ) D 1 2 for z , z ∈ D. 1 2 Problem 2.1 For z , z ∈ D, ﬁnd the point u ∈ ∂ D such that the sum |z −u|+|z −u| 1 2 1 2 is minimal. The point u is given as the point of tangency of an ellipse |z − z |+|z − z |= r 1 2 with the unit circle. Remark 2.2 For z , z ∈ D,if u ∈ ∂ D is the point of tangency of an ellipse |z − z |+ 1 2 1 |z − z |= r and the unit circle, then r is given by r =|2 − uz − uz |. 1 2 In fact, from the “reﬂective property” (z , u, 0) = (0, u, z ) of an ellipse, the 1 2 following holds u u − z u − z 2 2 arg = arg =− arg , (2.1) u − z u u and arg(u(u − z )) = arg(u(u − z )). (2.2) 1 2 Since the point u is on the ellipse |z − z |+|z − z |= r and satisﬁes uu = 1, we 1 2 have r =|u − z |+|u − z |=|u(u − z )|+|u(u − z )| 1 2 1 2 =|u(u − z ) + u(u − z )|=|2 − uz − uz |. 1 2 1 2 2.1 Proof of Theorem 1.1 From the Eq. (2.1), we have u − z u − z 1 2 arg · = 0. u u (u − z )(u − z ) 1 2 This implies is real and its complex conjugate is also real. Hence, (u − z )(u − z ) (u − z )(u − z ) 1 2 1 2 holds. Since u satisﬁes uu = 1, we have the assertion. 123 The Ptolemy–Alhazen Problem... 139 Remark 2.3 The solution of (1.3) includes all the tangent points of the ellipse |z − z |+|z − z |=|2 −uz −uz | and the unit circle. (See Figs. 1, 2). Figure 2 displays a 1 2 1 2 situation where all the roots of the quartic equation have unit modulus. However, this is not always the case for the Eq. (1.3). E.g., if z = 0.5+(0.1·k)i , k = 1, .., 5, z = 0.5, 1 2 the Eq. (1.3) has two roots of modulus equal to 1 and two roots off the unit circle, see Fig. 3. Miller and Vegh [18] computed the grazing angle of specular reﬂection using a quartic self-inversive polynomial equation, which is not the same as (1.3). Note that all the roots of their equation have modulus equal to one. They have also studied the Ptolemy–Alhazen problem using a quartic equation, that is different from our equation and, moreover, all the roots of their equation have modulus equal to one Fig. 3. We say that a polynomial P(z) is self-inversive if P(1/u) = 0 whenever u = 0 and P(u) = 0 . It is easily seen that the quartic polynomial in (1.3) is self-inversive. Note that the points u and 1/u are obtained from each other by the inversion transformation w → 1/w. It is clear from the compactness of the unit circle, that the function |z −z|+|z −z| 1 2 attains its maximum and minimum on the unit circle. However, as a property of the Eq. (1.3) itself, the following results can also be derived. Lemma 2.4 The Eq. (1.3) always has at least two roots of modulus equal to 1. Proof Consider ﬁrst the case, when z z = 0 . In this case, the Eq. (1.3) has two roots 1 2 u, |u|= 1, with u = z /z ∈ ∂ D if z = 0, z = 0 . (The case z = z = 0is 1 1 2 1 1 2 trivial.) Suppose that the equation has no root on the unit circle ∂ D . By the invariance property pointed out above, if u ∈ C \ ({0}∪ ∂ D) is a root of (1.3), then 1/u also is a root of (1.3). Hence, the number of roots off the unit circle is even and the number of roots on the unit circle must also be even. We will now show that this even number is either 2 or 4. Let a, b,α,β ∈ R, 0 < a < 1, 0 < b < 1 , and let 1 1 i α i α i β i β ae , e , be , e a b Fig. 1 Light reﬂection on a circular arc: The angles of incidence and reﬂection are equal. Ptolemy–Alhazen interior problem: Given z and z , ﬁnd 1 2 u. The maximal ellipse contained in the unit disk with foci z and z meets the unit 1 2 circle at u 123 140 M. Fujimura et al. Fig. 2 This ﬁgure indicates the four solutions of (1.3) (dots on the unit circle) and the ellipse that corresponds to each u,for z = 0.5 + 0.5i , z =−0.8i. The ﬁgure on the lower right shows the point u that gives the 1 2 minimum be the four roots of the Eq. (1.3). Then, the equation 1 1 i α i α i β i β z z (u − ae ) u − e (u − be ) u − e = 0 (2.3) 1 2 a b coincides with (1.3). Therefore, the coefﬁcient of degree two of (2.3) vanishes, and we have 1 1 i2α i2β i (α+β) e + e =− a + b + e . (2.4) a b The absolute value of the left hand side of (2.4) satisﬁes i2α i2β |e + e |≤ 2. (2.5) 123 The Ptolemy–Alhazen Problem... 141 Fig. 3 For z = 0.5 + 0.5i and z = 0.5, there are only two solutions of (1.3) on the unit circle. The ﬁgure 1 2 on the lower right shows the point u that gives the minimum On the other hand, the absolute value of the right hand side of (2.4) satisﬁes 1 1 1 1 i2(α+β) a + b + e = a + b + > 4, (2.6) a b a b because the function f (x ) = x + is monotonically decreasing on 0 < x ≤ 1 and f (1) = 2. The inequalities (2.5) and (2.6) imply that the equality (2.4) never holds. Hence (1.3) has roots of modulus equals to 1. Remark 2.5 We consider here several special cases of the Eq. (1.3) and for some special cases, we give the corresponding formula for the s metric which readily follows from Corollary 1.2. 123 142 M. Fujimura et al. Case 1. z = 0 = z (cubic equation). The Eq. (1.3)isnow (−z ) u + z u = 0 1 2 1 1 and has the roots u = 0, u =± and for z ∈ D 1 2,3 |z | |z| s (0, z) = . 2 −|z| Case 2. z + z = 0, z = 0. The Eq. (1.3) reduces now to: 1 2 1 2 4 z z 1 1 2 4 2 4 4 −z u + z = 0 ⇔ u = ⇔ u = . z |z | 1 1 z z 1 1 The roots are: u =± , u =±i (four distinct roots of modulus 1,2 3,4 |z | |z | 1 1 1) and for z ∈ D s (z, −z) =|z| . Case 3. z = z = 0 . Clearly s (z, z) = 0 . Denote z := z = z .The Eq.(1.3) 1 2 D 1 2 reduces now to: 2 4 3 2 2 2 z u − 2zu + 2zu − z = (zu − z)(zu − 2u + z) = 0 . Then, we see that u =± are roots. The other roots are: 1,2 |z| 1 2 1) If |z| < 1, then u = 1 ± 1 − |z| (with |u | > 1, |u | < 1) 3,4 3 4 1 2 2) If |z| > 1, then u = 1 ± i |z| − 1 (with |u | = |u | = 1). 3,4 3 4 Case 4. |z | = |z | = 0 . 1 2 Denote ρ = |z | = |z |. Using a rotation around the origin and a change of 1 2 orientation, we may assume that arg z =− arg z =: α, where 0 ≤ α ≤ . 2 1 2 4 3 2 The Eq. (1.3) reads now: ρ u − 2ρ cos α u + 2ρ cos α u − ρ = 0 ( ) ( ) 2 cos α 2 4 3 2 2 2 2 ρ u − 2ρ (cos α) u + 2ρ (cos α) u − ρ = ρ u − 1 u − u + 1 The roots are: u =±1 and 1,2 cos α cos α 1) If 0 <ρ < cos α, then u = ± − 1 (here |u | > 1, |u | < 1) 3,4 3 4 ρ ρ cos α cos α 2) If ρ ≥ cos α, then u = ± i 1 − (here |u | = |u | = 1). 3,4 3 4 ρ ρ Note that Case 4 includes Cases 2 and 3 (for α = , respectively, α = 0). Case 5. z = tz (t ∈ R, z = 0). This case is generalization of cases z = 0 = z , 1 2 2 1 2 z + z = 0, z = 0 and z = z = 0. 1 2 1 1 2 4 3 Denote P (u) = z z u − (z + z ) u + (z + z ) u − z z . 1 2 1 2 1 2 1 2 123 The Ptolemy–Alhazen Problem... 143 Denoting z = z we have: 2 4 3 2 P(u) = tz u − (1 + t ) zu + (1 + t ) zu − tz 4 2 z z 2 4 2 = tz u − − (1 + t ) zu u − . 4 2 |z| |z| z z P(u) = z u − u + tzu − (1 + t ) u + tz |z| |z| For t = 0 the roots of P are 0, ± . |z| Let t = 0. Besides ± there are two roots, which have modulus 1 if and |z| 1+t only if |z| ≥ . 2t 2.2 Exterior Problem Given z , z ∈ C \ D, ﬁnd the point u ∈ ∂ D such that the sum |z − u|+|z − u| is 1 2 1 2 minimal. Lemma 2.6 If the segment [z , z ] does not intersect with ∂ D, the point u is given as 1 2 a solution of the equation 4 3 z z u − (z + z )u + (z + z )u − z z = 0. 1 2 1 2 1 2 1 2 Remark 2.7 The above equation coincides with the Eq. (1.3) for the “interior problem”, since Theorem 1.1 could be proved without using the assumption z , z ∈ D. 1 2 Remark 2.8 The equation of the line joining two points z and z is given by 1 2 z − z z − z 1 1 = . (2.7) z − z z − z 2 2 Then, the distance from the origin to this line is |z z − z z | 1 2 1 2 2|z − z | 1 2 |z z − z z | 1 2 1 2 Therefore, if two points z , z satisfy ≤ 1, the line (2.7) intersects with 1 2 2|z − z | 1 2 the unit circle, and the triangular ratio metric s (z , z ) = 1. 1 2 C\D Lemma 2.9 The boundary of B (z, t ) ={w ∈ D : s (z,w) < t } is included in an s D algebraic curve. Proof Without loss of generality, we may assume that the center point z =: c is on the positive real axis. Then, 123 144 M. Fujimura et al. |c − w| s (c,w) = sup |c − ζ|+|ζ − w| ζ ∈∂ D |c − w| = (from Remark 2.2), (2.8) |2 − uc − uw| where u is a minimizing root of the equation 4 3 U (w) = cwu − (c + w)u + (c + w)u − cw = 0. (2.9) 2t Moreover, B (0, t ) ={|w| < } (resp. B (c, 0) ={c}) holds for c = 0(resp. s s 1+t t = 0), and B (c, t ) ={0} holds if and only if c = 0 and t = 0. Therefore, we may assume that c = 0, t = 0 and w ≡ 0. Now, consider the following system of equations s (c,w) = t and U (w) = 0, i.e, D c 2 2 2 S (w) = t |2 − uc − uw| −|c − w| = 0 and U (w) = 0. (2.10) c,t c The above two equations have a common root if and only if both of the polynomials S (w) and U (w) have non-zero leading coefﬁcient with respect to u variable and c,t c the resultant satisﬁes resultant (S , U ) = 0. Using the “resultant” command of the u c,t c Risa/Asir software, we have resultant (S , U ) = cww · B (w), u c,t c c,t where B (w) c,t 2 2 8 = (wc − 1)(wc − 1) (c + ww − 2) − 4(wc − 1)(wc − 1) t 8 7 2 2 6 − (c − w)(c − w) 4wwc − 3(w + w)c − 2(2w w + 2ww − 1)c 5 3 3 2 2 − (w + w)(13ww + 2)c − 2(2w w − (36w + 10)w − 27ww 2 4 2 2 3 − 10w − 4)c − (w + w)(13w w + 92ww + 32)c 3 3 2 2 2 2 2 + 2(ww(2w w − 2w w + 27ww + 48) + 2(5ww + 2)(w + w ))c 2 2 2 2 6 − ww(w + w)(3w w + 2ww + 32)c + 2w w (ww + 4) t 2 2 6 5 2 2 4 + (c − w) (c − w) 6wwc − 3(w + w)c + (4w w + 16ww + 1)c 3 3 3 2 2 2 2 − 2(w + w)(13ww + 5)c + (6w w + (16w + 1)w + 52ww + w )c 2 2 4 − ww(w + w)(3ww + 10)c + w w t 3 3 2 2 2 − c(c − w) (c − w) 4wwc(c + ww + 3) − (c + ww)(w + w) t 2 4 4 + c ww(c − w) (c − w) . Moreover, we can check that 2 2 8 B (w) =|w| c |c − w| c,0 123 The Ptolemy–Alhazen Problem... 145 and 4 4 2 2 2 2 2 2 B (w) =|w| t (t − 1) |w| − 4t (t + 1) |w| − 4t . 0,t Hence, the boundary of B (c, t ) is included in the algebraic curve deﬁned by the equation B (w) = 0. c,t Remark 2.10 The algebraic curve {w : B(w) = 0} does not coincide with the boundary ∂ B (c, t ). There is an “extra” part of the curve since the Eq. (2.9) contains extraneous solutions. The analytic formula in Corollary 1.2 for the triangular ratio metric s (z , z ) is D 1 2 not very practical. Therefore, we next give an algorithm based on Theorem 1.1 for the evaluation of the numerical values. Algorithm. We next give a Mathematica algorithm for computing s (x , y) for given points x , y ∈ D. Figure 4 was drawn with the help of this algorithm. sD[x_, y_] := Module[{u, sol, mySol, tmp = 2*Sqrt[2]}, sol = Solve[ Conjugate[ x*y] uˆ4 - Conjugate[x + y] uˆ3 + (x + y) u - x*y == 0, {u}]; mySol = u /. sol; Do[If[Abs[Abs[mySol[[i]] ] - 1] < 10ˆ(-12), tmp = Min[tmp, Abs[mySol[[i]] - x] + Abs[mySol[[i]] - y]]], {i, 1, Length[mySol]}]; Abs[x - y]/tmp] ; One can also use numerical methods to compute s , see [6]. Fig. 4 Level sets {x + iy : s (0.3, x + iy) = t } for t = 0.1, 0.2, 0.3, 0.4, 0.6 0.6 and the unit circle. By Lemma 2.9, these level sets are contained in an algebraic curve. These level sets are drawn with the help of the Mathematica algorithm below 0.3 0.1 0.2 0.4 123 146 M. Fujimura et al. 3 Geometric Approach to the Ptolemy–Alhazen Problem In this section, the unimodular roots of Eq. (1.3) are characterized as points of intersec- tion of a conic section and the unit circle, then n such roots are studied, where n = 4 in the case of the exterior problem and n = 2 in the case of the interior problem. We describe the construction of the conic section mentioned above. Except in the cases where 0, z , z are collinear or |z | = |z | , the construction cannot be carried out as 1 2 1 2 ruler-and-compass construction. Neumann [19] proved that Alhazen’s interior problem for points z , z is solvable by ruler and compass only for (Rez , Imz , Rez , Imz ) 1 2 1 1 2 2 belonging to a null subset of R , in the sense of Lebesgue measure. We characterize algebraically condition (1.1) without assuming that z , z ∈ D,or 1 2 z , z ∈ C \ D,or u ∈ ∂ D. 1 2 Lemma 3.1 Let z , z ∈ C and u ∈ C \ {z : k = 1, 2}. The following are equivalent: 1 2 k (i) (z , u, 0) = (0, u, z ). 1 2 2 2 2 2 u u u u (ii) = and + > 0; (u−z )(u−z ) (u−z )(u−z ) (u−z )(u−z ) (u−z )(u−z ) 1 2 1 2 1 2 1 2 (iii) 2 2 2 2 z z u − (z + z ) uu + (z + z ) u u − z z u = 0 (3.1) 1 2 1 2 1 2 1 2 and 2 2 2 2 2 2 z z u − (z + z ) uu − (z + z ) u u + z z u + 2u u > 0. (3.2) 1 2 1 2 1 2 1 2 Proof Let u ∈ C \ {z : k = 1, 2}. Clearly, (z , u, 0) = arg and (0, u, z ) = k 1 2 u−z u−z u−z 2 u 2 arg . Denoting v := : , we see that (z , u, 0) = (0, u, z ) if and 1 2 u u−z u only if v satisﬁes both v = v and v + v> 0, i.e. if and only if (ii) holds. We have v = v (respectively, v + v> 0) if and only if (3.1) (respectively, (3.2)) holds, therefore (ii) and (iii) are equivalent. In the special case z = z = 0(z = z = 0) (i), (ii) and (iii) are satisﬁed whenever 1 2 1 2 u ∈ C (respectively, if and only if u = λz for some real number λ = 0, 1). Remark 3.2 Let u ∈ C \ {z : k = 1, 2} . If 2 2 2 u u u = and (u − z )(u − z ) (u − z )(u − z ) (u − z )(u − z ) 1 2 1 2 1 2 + < 0, u − z u − z ( )( ) 1 2 then |(z , u, 0) − (0, u, z )| = π. The converse also holds. 1 2 Consider the interior problem, with z , z ∈ D and u ∈ ∂ D. The unit circle is 1 2 exterior to the circles of diameters [0, z ], [0, z ]. An elementary geometric argu- 1 2 π π π π ment shows that − < (z , u, 0)< and − < (0, u, z )< , therefore 1 2 2 2 2 2 |(z , u, 0) − (0, u, z )| = π. In this case (3.1) implies (z , u, 0) = (0, u, z ). 1 2 1 2 The Eq. (3.1) deﬁnes a curve passing through 0, z and z , that is a cubic if z +z = 1 2 1 2 0, respectively, a conic section if z + z = 0 with z , z ∈ C . Then, under the 1 2 1 2 123 The Ptolemy–Alhazen Problem... 147 Fig. 5 The exterior problem. Intersection of the conic (3.3) with the unit circle inversion with respect to the unit circle, the image of the curve given by (3.1) has the equation 2 2 z z u − (z + z ) u + (z + z ) u − z z u = 0. (3.3) 1 2 1 2 1 2 1 2 This is a conic section, that degenerates to a line if z z = 0 with z , z not both zero. 1 2 1 2 The points of intersection of the unit circle with the conic section (3.3)are shownin Fig. 5. Remark 3.3 If u ∈ ∂ D, then (3.1) (respectively, (3.3)) holds if and only if 1 1 z z u − (z + z ) u + (z + z ) − z z = 0. 1 2 1 2 1 2 1 2 u u The Eqs. (3.3), (3.1) and (1.3) have the same unimodular roots. Lemma 3.4 Let z , z ∈ C . The conic section given by (3.3) has the center c = 1 2 1 1 1 1 1 1 1 + and it passes through 0, , , + .If |z | = |z | or |arg z − arg z | ∈ 1 2 1 2 2 z z z z z z 1 2 1 2 1 2 {0,π }, then consists of the parallels d ,d through c to the bisectors (interior, 1 2 respectively, exterior) of the angle (z , 0, z ). In the other cases, is an equilateral 1 2 hyperbola having the asymptotes d and d . 1 2 Proof The Eq. (3.3) is equivalent to 1 1 Im z z u + − u = 0. (3.4) 1 2 z z 1 2 1 1 The curve passes through the points 0 and 2c = + .If u satisﬁes (3.4), then z z 1 2 2c − u also satisﬁes (3.4); therefore, has the center c. Since z and z are on the 1 2 1 1 cubic curve given by (3.1), passes through and . The conic section is a pair z z 1 2 1 1 1 of lines if and only if passes through its center. For u = + we have 2 z z 1 2 1 1 1 z z 1 2 Im z z u + − u = Im + , 1 2 z z 4 z z 1 2 2 1 123 148 M. Fujimura et al. z z 1 2 therefore is a pair of lines if and only if + ∈ R. The following conditions are z z 2 1 equivalent: z z 1 2 (1) + ∈ R; z z 2 1 z z 2 2 (2) ∈ R or = 1; z z 1 1 | | { } | | | | (3) arg z − arg z ∈ 0,π or z = z . 1 2 1 2 Denote u = x +iy. Using a rotation around the origin and a reﬂection we may assume that arg z =− arg z =: α, where 0 ≤ α ≤ . In this case, the equation of is 2 1 2 2 |z | + |z | |z | − |z | |z | − |z | 1 2 2 1 2 1 x − cos α y − sin α = sin 2α. (3.5) 2 |z z | 2 |z z | 8 |z z | 1 2 1 2 1 2 The Eq. (3.5) shows that is the pair of lines d , d if |z | = |z | or sin 2α = 0; 1 2 1 2 otherwise, is an equilateral hyperbola having the asymptotes d and d . 1 2 Lemma 3.5 (Sylvester’s theorem) In any triangle with vertices z , z , z , the ortho- 1 2 3 center z and the circumcenter z satisfy the identity z + 2z = z + z + z . H C H C 1 2 3 z +z +z 1 2 3 Proof Let z be the centroid of the triangle. It is well known that z = . G G By Euler’s straightline theorem, z − z = 2(z − z ). Then z + 2z = 3z = H G G C H C G z + z + z . 1 2 3 1 1 Lemma 3.6 Let z , z ∈ C . The orthocenter of the triangle with vertices 0, , 1 2 z z 1 2 belongs to the conic section given by Eq. (3.3). Proof Consider a triangle with vertices z , z , z and denote by z and z the 1 2 3 H C orthocenter and the circumcenter, respectively. By Sylvester’s theorem, Lemma 3.5, z = z + z + z − 2z . H 1 2 3 C ⎛ ⎞ ⎛ ⎞ But 111 111 ⎝ ⎠ ⎝ ⎠ z = det z z z : det z z z . C 1 2 3 1 2 3 2 2 2 |z | |z | |z | z z z 1 2 3 1 2 3 z z (z −z ) 1 2 2 1 If z = 0, then z = , hence 3 C z z −z z 1 2 1 2 (z − z )(z z + z z ) 1 2 1 2 1 2 z = . z z − z z 1 2 1 2 1 1 Let h be the orthocenter of the triangle with vertices 0, , . The above formula z z 1 2 implies z − z z z + z z 2 1 1 2 1 2 h = . (3.6) z z z z − z z 1 2 1 2 1 2 2 2 Let f (u) := z z u − (z + z ) u + (z + z ) u − z z u . Then f (u) = 2iIm 1 2 1 2 1 2 1 2 2z z (z −z ) 2 1 2 2 1 z z u − (z + z ) u . Since z z h − (z + z ) = , it follows that 1 2 1 2 1 2 1 2 z z −z z 1 2 1 2 −16 |z − z | 2 1 2 2 z z h − (z + z ) h = Re (z z ) Im (z z ) 1 2 1 2 1 2 1 2 |z z − z z | 1 2 1 2 is a real number, hence f (h) = 0. 123 The Ptolemy–Alhazen Problem... 149 Let z , z ∈ C be such that |z | = |z | and |arg z − arg z | ∈ / {0,π }.Let h be 1 2 1 2 1 2 2(z −z ) 1 1 2 1 1 1 given by (3.6). Note that h − + = = 0. If h ∈ / 0, , then the z z z z −z z z z 1 2 1 2 1 2 1 2 1 1 1 1 hyperbola passing through the ﬁve points 0, , , + , h can be constructed z z z z 1 2 1 2 using a mathematical software. 1 1 In the cases where h ∈ 0, , , we choose a vertex of the hyperbola as the z z 1 2 ﬁfth point needed to construct . The vertices of the equilateral hyperbola are the intersections of with the line passing through the center of the hyperbola, with the arg z −arg z 2 1 slope m = 1if |z | > |z |, respectively, m =−1if |z | < |z |.Let α := . 1 2 1 2 Using (3.5) it follows that the distance d between a vertex and the center of is 2 2 | | | | √ z − z 1 2 d = sin 2α. 2|z z | 1 2 2 2 π 1 1 1 1 If h = 0we have α = and d = − . Assume that h = , the case 4 2 z z z 2 1 1 2 2 2 h = being similar. Then |z | = |z | cos 2α< |z | and |z | − |z | = |z − z | , 2 1 1 1 2 1 2 1 1 1 1 therefore d = − sin 2α.Let z be the orthogonal projection of on the 2 z z z 2 1 1 1 1 1 1 1 line joining to the origin. Then d = − · − z . We see that a vertex z 2 z z z 2 2 1 2 1 1 of can be constructed with ruler and compass if h ∈ 0, , . z z 1 2 1 1 Remark 3.7 Being symmetric with respect to the center of , and belong to z z 1 2 1 1 distinct branches of , each branch being divided by or into two arcs. If z ∈ C\D, z z 1 2 k ∈ {1, 2}, then each of these arcs joins , that is in the unit disk, with some point exterior to the unit disk; therefore, it intersects the unit circle. It follows that, in the case of the exterior problem, intersects the unit circle at four distinct points. In the following, we identify the points of intersection of the conic section given by (3.3) with the unit circle. After ﬁnding the points u ∈ ∂ D ∩ , it is easy to select among these the points u for which (1.1) holds, respectively, for which |u − z | + |u − z | 1 2 attains its minimum or its maximum on ∂ D. First assume that is a pair of lines d , d , parallel to the interior bisector and to the 1 2 exterior bisector of the angle (z , 0, z ), respectively. Let α = |arg z − arg z |. 1 2 2 1 Then α ∈ 0, or |z | = |z |. The distances from the origin to d and d are 1 2 1 2 ||z |−|z || |z |+|z | 2 1 1 2 δ = sin α and δ = cos α. Then, intersects the unit circle at 1 2 2|z z | 2|z z | 1 2 1 2 four distinct points in the following cases: (i) z , z ∈ C \ D; (ii) z , z ∈ D with 1 2 1 2 1 1 1 − < 1 or with |z | = |z | > cos α. In the other cases for z , z ∈ D the 1 2 1 2 2 |z | |z | 1 2 intersection of with the unit circle consists of two distinct points. Proposition 3.8 If the conic section given by (3.3) is a hyperbola, then the intersec- tion of with the unit circle consists of (i) four distinct points if z , z ∈ C \ D, one in the interior of each angle determined 1 2 by the lines that pass through the origin and z , respectively, z ; 1 2 (ii) at least two distinct points if z , z ∈ D, one in the interior of the angle determined 1 2 by the rays passing starting at the origin and passing through z , respectively, z 1 2 and the other in the interior of the opposite angle. 123 150 M. Fujimura et al. it Proof The intersection of with the unit circle consists of the points u = e , t ∈ (−π, π ] satisfying i2t −it Im z z e − (z + z ) e = 0. 1 2 1 2 Let z , z ∈ C . There are at most four points of intersection of and the unit circle, 1 2 since these are the roots of the quartic Eq. (1.3). Using a rotation around the origin and a change of orientation, we may assume that arg z =− arg z =: α, where 0 ≤ α ≤ . The above equation is equivalent to 2 1 g (t ) := |z z | sin 2t − |z | sin (t + α) − |z | sin (t − α) = 0. (3.7) 1 2 1 2 We have g (−π ) = g (π ) =−g (0) = (|z | − |z |) sin α, 1 2 g (α − π ) = |z | (|z | + 1) sin 2α, g (−α) = |z | (1 − |z |) sin 2α, 1 2 2 1 g (α) = |z | (|z | − 1) sin 2α, g (π − α) =− |z | (|z | + 1) sin 2α. 1 2 2 1 Consider the cases where is a hyperbola, i.e., 0 <α < . Clearly, −π< α − π< −α< 0 <α <π − α< π.Wehave g (π − α) < 0 < g (α − π ), while g (−π ) = g (π ) =−g (0) has the same sign as |z | − |z |. 1 2 (i) Assume that z , z ∈ C \ D. Then g (−α) < 0 and g (α) > 0. 1 2 If |z | < |z |, then g (−π ) < 0 < g (α − π ) > 0, g (−α) < 0 < g (0) and 1 2 g(α) > 0 > g (π − α). Since g is continuous on R,Eq. (3.7) has at least one root in each of the open intervals (−π, α − π ), (α − π, −α), (−α, 0) and (α, π − α). If |z | < |z |, then g (α − π ) > 0 > g (−α), g (0) < 0 < g(α) and g (π − α) < 2 1 0 < g (π ).The Eq.(3.7) has at least one root in each of the open intervals (α − π, −α), (0,α), (α, π − α) and (π − α, π ). (ii) Now assume that z , z ∈ D. Then g (−α) > 0 and g (α) < 0. 1 2 If |z | < |z |, then g (−π ) < 0 < g (α − π ) and g (0) > 0 > g(α). Since g 1 2 is continuous on R,Eq. (3.7) has at least one root in each of the open intervals (−π, α − π ) and (0,α). | | | | If z > z , then g (0) > 0 > g(α) and g (π − α) < 0 < g (π ).The Eq.(3.7) 1 2 has at least one root in each of the open intervals 0,α and π − α, π . ( ) ( ) Corollary 3.9 The Eq. (1.3) has four distinct unimodular roots in the case of the exterior problem and has at least two distinct unimodular roots in the case of the interior problem. 4 Remarks on the Roots of the Equation (1.3) In this section, we study the number of the unimodular roots of the Eq. (1.3) (i.e., the roots lying on the unit circle) and their multiplicities. Denote P(u) = z z u − 1 2 123 The Ptolemy–Alhazen Problem... 151 (z + z )u + (z + z )u − z z . If either z = 0or z = 0 then the cubic Eq. (1.3) 1 2 1 2 1 2 1 2 P (u) = 0 has a root u = 0 and two simple roots on the unit circle. We will assume in the following that z = 0 and z = 0. As we observed in Sect. 2, 1 2 the quartic polynomial P is self-inversive. Then, P has an even number of zeros on the unit circle, each zero being counted as many times as its multiplicity. According to Lemma 2.4, P has at least two unimodular zeros, distinct or not, that is P has four or two unimodular zeros. There is a rich literature dealing with the location of zeros of a complex self-inversive polynomial with respect to the unit circle [5,7,8,15–17]. 4 3 Lemma 4.1 P(u) = z z u − (z + z )u + (z + z )u − z z cannot have two double 1 2 1 2 1 2 1 2 zeros on the unit circle. Proof Assume that P has two double zeros a and b on the unit circle, P(u) = z z (z − 1 2 2 2 2 a) (z − b) (a, b ∈ ∂ D, a = b). Since the coefﬁcient of u in P (u) vanishes, 2 2 a + 4ab + b = a + (2 − 3)b a + (2 + 3)b = 0 . This contradicts the assumption |a|=|b|= 1. Similarly, we rule out another case. 4 3 Lemma 4.2 For P(u) = z z u − (z + z )u + (z + z )u − z z it is not possible 1 2 1 2 1 2 1 2 to have a double zero on the unit circle and two zeros not on the unit circle. Proof Assume that P has a double zero a with |a|= 1 and the zeros b = . Then 2 1 2 P(u) = z z (z − a) (z − b) z − . The coefﬁcient of u in P (u) vanishes, 1 2 b 1 a + + 2a b + = 0. b b We have 1 1 1 1 | | b + = b + b + = 2 + b + > 4. b |b| b b b 1 Then 2 ≥ a + = 2a b + > 4, a contradiction. b b 4 3 Lemma 4.3 If P(u) = z z u − (z + z )u + (z + z )u − z z has a triple zero 1 2 1 2 1 2 1 2 a and a simple zero b, then b =−a, with a and b lying on the unit circle and |z + z | = 2 |z z |. 1 2 1 2 Proof Assume that P has a triple zero a and a simple zero b, P(u) = z z (z − 1 2 a) (z − b), where a, b ∈ C, a = b. Since P is self-inversive, |a|=|b|= 1 and b = =−a. Also, the fact that the coefﬁcient of u in P(u) vanishes already implies a(a + b) = 0. But z z a b =−z z = 0, therefore b =−a. Considering the 1 2 1 2 3 3 coefﬁcient of u in P(u) = z z (u − a) (u + a),itfollows that 2az z = z + z , 1 2 1 2 1 2 hence |z + z | = 2 |z z |. 1 2 1 2 123 152 M. Fujimura et al. 4 3 Example 4.4 Find the relation between z , z such that P(u) = z z u −(z +z )u + 1 2 1 2 1 2 (z + z )u − z z has the triple zero 1 and the simple zero (−1). 1 2 1 2 Suppose P(u) = z z (u − 1) (u + 1) = 0 . (4.1) 1 2 From the constant term of (1.3) and (4.1), we have z z ∈ R. Similarly, from the 1 2 coefﬁcient of u in (1.3) and (4.1), we have z + z − 2z z = 0 . 1 2 1 2 Therefore z and z coincide with the two solutions of w − 2 pw + p = 0, where 1 2 p = z z ∈ R (in particular −1 < p < 1 for the interior problem). 1 2 In the case where 0 < p < 1, z and z are complex conjugates to each other since 1 2 2 2 3 discriminant(w − 2 pw + p,w) = 4(p − p)< 0. Hence, P(u) = z z (u − 1) (u + 1 1 1) = 0, and we have 1 1 (2z − 1)z − z = 2 z − − = 0. 1 1 1 1 2 2 1 1 Therefore, for z on the circle |z − |= and z = z , P(u) = 0 has exactly two 1 2 1 2 2 roots 1 and −1. This case was studied in [11, Thm. 3.1]. In fact, for z = a + bi with 2 2 3 a − a + b = 0, P(u) = a(u − 1) (u + 1) = 0. In the case where −1 < p < 0, the quadratic equation w − 2 pw + p = 0 has two real roots and we have P(u) = z z (u − 1) (u + 1). 1 2 Moreover, we can parametrize two foci as follows, z = t , z = (−1 < t < 1 2 2t −1 2 − 1). It remains to study the following cases: Case 1. P has four simple unimodular zeros. Case 2. P has two simple unimodular zeros and two zeros that are not unimodular. Case 3. P has a double unimodular zero and two simple unimodular zeros. ∗ 4 3 Proposition 4.5 Assume that z , z ∈ C . Let P(u) = z z u − (z + z )u + (z + 1 2 1 2 1 2 1 z )u − z z . Then 2 1 2 a) P has four simple unimodular zeros if |z + z | < |z z | and 1 2 1 2 b) P has exactly two unimodular zeros, that are simple, if |z + z | > 2 |z z |. 1 2 1 2 c) If P has four simple unimodular zeros, then |z + z | < 2 |z z |. 1 2 1 2 d) If P has exactly two unimodular zeros, that are simple, then |z + z | > |z z |. 1 2 1 2 Proof Let f be a complex polynomial. The location of the zeros of the derivative f of f is connected with the location of the zeros of f . Gauss–Lucas theorem [17, Thm. 6.1] shows that the zeros of the derivative f lie within the convex hull of the set of zeros of f . In particular, if all the zeros of f lie on the unit circle, then all the zeros of f lie in the closed unit disk (and f is self-inversive). The converse holds by a theorem of Cohn [8] stating that a complex polynomial has all its zeros on the unit 123 The Ptolemy–Alhazen Problem... 153 circle if and only if the polynomial is self-inversive and its derivative has all its zeros in the closed unit disk. 3 2 2 In our case P (u) = 4z z u −3 (z + z ) u +(z + z ) and P (u) = 12z z u − 1 2 1 2 1 2 1 2 6 (z + z ) u. 1 2 a) Assume that |z + z | < |z z |. Then for u ∈ ∂ D we have 1 2 1 2 3 2 4z z u = 4 |z z | > 4 |z + z | ≥ −3 (z + z ) u + (z + z ) 1 2 1 2 1 2 1 2 1 2 It follows by Rouché’s theorem [22, 3.10] that the derivative P has all its zeros in the unit disk. By Cohn’s theorem cited above, P has all its four zeros on the unit circle ∂ D. Assume that the polynomial f is self-inversive. By [7, Thm.1], the following are equivalent: (i) all the zeros of f are simple and unimodular; (ii) there exist a polynomial g having all its zeros in the unit disk |z| < 1, a non- m i θ ∗ negative integer m and a real number θ such that f (z) = z g (z) + e g (z) ∗ n 1 for all z ∈ C.Here g (z) := z g , where n = deg Q. m i θ ∗ In our case, P (u) = u Q (z) + e Q (u) for m = 3, θ = π and Q (u) = | | 3 3 z +z 1 2 z z u + (z + z ). The roots of Q have modulus < 1. The implication 1 2 1 2 |z z | 1 2 (ii ) ⇒ (i ) from [7, Thm. 1] shows that P has four simple zeros on the unit circle. b) Now assume that |z + z | > 2 |z z |.For u ∈ ∂ D we have 1 2 1 2 2 3 −3 (z + z ) u = 3 |z + z | > 4 |z z | + |z + z | ≥ 4z z u + (z + z ) 1 2 1 2 1 2 1 2 1 2 1 2 and it follows using Rouché’s theorem that P has exactly two zeros in the closed unit disk. Cohn’s theorem shows that P cannot have all its zeros on ∂ D.By Lemma 2.4, P has at least two unimodular zeros; therefore, P has exactly two unimodular zeros. By Lemma 4.2, these unimodular zeros are simple. An alternative way to prove that P has exactly two unimodular zeros is indicated below. Assume by contrary that P has four unimodular zeros. Using the Gauss– Lucas theorem two times, it follows that each of the derivatives P and P has all z +z 1 2 its zeros in the closed unit disc |z|≤ 1. The zeros of P are 0 and . Then, 2z z 1 2 under the assumption |z + z | > 2 |z z |, the second derivative P has a zero in 1 2 1 2 |z| > 1, which is a contradiction. c) Assume that P has four simple unimodular zeros. Then, P has all its zeros in the closed unit disk. Given a self-inversive polynomial f ,itisprovedin[5,Lem.] that each unimodular zero of the derivative f is also a zero of f .If P has a unimodular zero a, then P a = 0; therefore, a is a zero of P of multiplicity ( ) at least 2, a contradiction. It follows that P has all its zeros in the unit disk. By Gauss–Lucas theorem, the second derivative P also has all its zeros in the unit disk; therefore, |z + z | < 2 |z z |. 1 2 1 2 d) Now suppose that P has exactly two simple unimodular zeros, a and b.Let c and the other zeros of P, with |c| < 1. Then P (u) = z z (u − a)(u − b)(u − c) 1 2 123 154 M. Fujimura et al. u − . The coefﬁcient of u in P (u) vanishes; therefore, c 1 ab + + (a + b) c + = 0, c c |c| ab c ab 1 1 1 and a + b =− − . Because = < and < ,we 1 2 1 2 2 2 c+ |c| +1 c+ |c| +1 c+ c c z +z 3 1 2 get |a + b| < 1. Considering the coefﬁcient of u in P (u) we obtain = z z 1 2 1 |z +z | 1 1 2 | | a + b + c + . Then ≥ c + − a + b > 1. c |z z | c 1 2 i α i (α+t ) Example 4.6 Let z = (1 + t ) e and z = (1 + t )e , where t > 0 and α ∈ 1 2 (−π, π ]. By Corollary 3.9,the Eq.(1.3) has four simple unimodular roots in this case. |z +z | 1 2 −it On the other hand, = 1 + t 1 + e → 2as t → 0, therefore the constant ( ) |z z | 1 2 2 in Proposition 4.5 c) cannot be replaced by a smaller constant. We give a direct proof for the following consequence of Proposition 4.5. 4 3 Corollary 4.7 If P(u) = z z u − (z + z )u + (z + z )u − z z has one double 1 2 1 2 1 2 1 2 zero and two simple zeros on the unit circle, then |z z | ≤ |z + z | ≤ 2 |z z |. 1 2 1 2 1 2 Proof Assume that P has one double unimodular zero a and two simple unimodular zeros b, c. Then P (u) = z z (z − a) (z − b)(z − c). 1 2 The coefﬁcient of u in P (u) vanishes, a + bc + 2a (b + c) = 0. 3 z +z a +bc 1 2 Considering the coefﬁcient of u in P (u) we obtain = 2a+b+c = 2a− = z z 2a 1 2 3a −bc 3 bc |z +z | 3 bc |z +z | 3 bc 1 2 1 2 = a − . Then ≤ a + − = 2 and ≥ a − − = 2a 2 2a |z z | 2 2a |z z | 2 2a 1 2 1 2 Acknowledgements Open access funding provided by University of Turku (UTU) including Turku Uni- versity Central Hospital. This research was begun during the Romanian-Finnish Seminar in Bucharest, Romania, June 20–24, 2016, where the authors P.H., M.M., and M.V. met. During a workshop at the Tohoku University, Sendai, Japan, in August 2016 organized by Prof. T. Sugawa, M.F., P.H., and M.V. met and had several discussions about the topic of this paper. P.H. and M.V. are indebted to Prof. Sugawa for his kind and hospitable arrangements during their visit. This work was partially supported by JSPS KAKENHI Grant Number 15K04943. The second author was supported by University of Turku Foundation and CIMO. The authors are indebted to Prof. G.D. Anderson for a number of remarks on this paper and to the referee for several useful remarks and comments. 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